The Dominion of Dukes
\(\)For a quick introduction to the Duke and the game Dominion, click show more.
Every time the Duke makes an appearance in Dominion, I try to justify putting him in my deck. But in the hubbub of a live game, I never find time to work out precisely how many Dukes and Duchies I’d need to beat someone going for Provinces, or what distribution would best serve me.
Let’s use a little math and figure it out.
We’re going to determine:
- The ideal balance of Dukes and Duchies given a fixed number of $5-purchases.
- How many victory points that balance yields, so we can compare this strategy against others.
The Duke-Duchy Balance
We could brute force it easily, but where’s the fun in that? Instead, we’re going to do some simple calculus.
Each Duchy is worth \(3\) victory points, and each Duke is worth \(1\) victory point per Duchy. We’re trying to maximize victory points \(V\), under the constraint of a fixed number of $5-purchases \(p\).
For \(k = \) the number of Dukes and \(c = \) the number of Duchies, $$ V = 3c + kc $$ and the constraint is $$ p = k + c. $$ We’ve fixed \(p\) because we assume we have a rough idea how many $5-purchase opportunities we’re going to get before the game ends — we’re just deciding what ratio of Dukes to Duchies we’ll buy with those opportunities. Let’s now maximize \(V\)! \begin{align} k & = p – c \\ V & = 3c + (p – c)c \\ V & = 3c + pc – c^2 \end{align} For fixed \(p\), \(V\) is maximized at \begin{align} \frac{\partial V}{\partial c} & = 3 + p – 2c = 0 \\ c & = \frac{3 + p}{2}. \end{align} That’s fine for odd \(p\), but for even \(p\), the maximal \(c\) is a non-integer, so it isn’t a possible quantity of Duchy cards. We’ll want one of the two nearest integer points, $$c_\pm = \frac{3 + p}{2} \pm \frac{1}{2}$$ but which one?
Fortunately, \(V\) is a parabola, symmetric around its maximum \(c = \frac{3 + p}{2}\). Since both points are off by the same amount, \(\frac{1}{2}\), both points have the same value — they both maximize \(V\).
So, for \(p\) odd, \begin{align} c & = \frac{p+3}{2} \\ \end{align} and for \(p\) even, \begin{align} c_\pm & = \frac{p+3}{2} \pm \frac{1}{2} \text{ (both work)}. \end{align}
Let’s finish up our answer by putting it in terms of Dukes and Duchies.
For \(p\) odd, \begin{align} c & = \frac{c+k+3}{2} \\ 2c & = c + k + 3 \\ \\ c & = k + 3 \\ \end{align} And for \(p\) even, \begin{align} c_\pm & = \frac{c_\pm + k_\pm + 3}{2} \pm \frac{1}{2} \\ 2c_\pm & = c_\pm + k_\pm + 3 \pm 1 \\ \\ c_\pm & = k_\pm + 3 \pm 1 \\ \end{align} So, for \(p\) odd we want \(3\) more Duchies than Dukes, and for \(p\) even we want as close to that as possible.
A discrete framing
We can motivate this result with a discrete framing of the problem:
When you buy a Duke, you get \(c\) victory points.
When you buy a Duchy, you get \(k + 3\) victory points.
So, when \(c > k + 3\), you’re better off buying a Duke.
When \(c < k + 3\), you're better off buying a Duchy.
And, naturally, when \(c = k + 3\), you can buy either one.
This implies that, once you have your first \(3\) Duchies, you’re best off with $$ c = k + 3 $$ or as close to that as you can get.
Maximal victory points
How many victory points does the ideal Duke-Duchy balance yield?
For \(p\) odd, \begin{align} c & = k + 3 \\ \\ V & = 3c + kc \\ V & = c(k+3) \\ V & = c^2 \\ \end{align} and for \(p\) even, \begin{align} c_\pm & = k_\pm + 3 \pm 1 \\ \\ c_+ & = k_+ + 4 \\ V & = c_+(3+k_+)\\ V & = c_+(c_+ -1 ).\\ \\ c_- & = k_- + 2\\ V & = c_-(3+k_-)\\ V & = c_-(c_- + 1).\\ \end{align} Note that because \(c_+ = c_- + 1\), we see explicitly that \(V\) is the same for \(c_+\) and \(c_-\). \begin{align} V = c_\pm (c_\pm \mp 1) \end{align}
That’s nice, but it’s strategically useful to have this in terms of \(p\), as a good player can tell roughly how many $5-purchases she’s going to get before the game ends.
For odd \(p\), \begin{align} c & = k + 3 \\ k & = p – c \\ c &= p – c + 3 \\ 2c &= (p+3) \\ c &= (p+3)/2 \\ V &= \frac{(p+3)^2}{4} \end{align} and for even \(p\), \begin{align} c_\pm & = k_\pm + 3 \pm 1 \\ k_\pm & = p – c_\pm \\ c_\pm & = p – c_\pm + 3 \pm 1 \\ 2c_\pm & = p + 3 \pm 1 \\ c_\pm & = \frac{p + 3 \pm 1}{2} \\ \\ V & = \frac{p + 3 \pm 1}{2} \left( \frac{p+3 \pm 1}{2} \mp 1 \right) \\ V & = \frac{(p + 3 \pm 1) (p+3 \mp 1)}{4} \\ V & = \frac{(p + 4)(p+2)}{4}. \\ \end{align}
Let’s use these formulas to make a graph for quick reference during a game:
In a game where everyone’s doing about the same, the first player to \(4\) Provinces usually wins. That’s worth \(24\) victory points, so we see that you need \(7\) $5-purchases to beat that player (barely).
Even if your opponents can get more than \(4\) Provinces (after all, there are more available given that you’re not buying any), once you reach \(p=7\) you’ll keep pace with your Province-purchasing opponents if you simply match them buy-for-buy, pulling slightly ahead of them again at \(p=11\) when they buy their \(8^{\text{th}}\) Province.
It’s noteworthy that because you need more Duchies than Dukes, you should bias yourself in the early game toward Duchies over Dukes beyond the \(c=k+3\) that maximizes victory points. There are two reasons for this:
- If an opponent decides to do the Duke-Duchy strategy, Duchies are going to run out before Dukes.
- Province-purchasers will snap up the occasional Duchy on their own, both for the VPs and to keep it out of your Duke deck. This is especially true in 2-player, where denying an opponent the ability to score is just as valuable as scoring yourself.
TL;DR
In summary, the ideal Duke-Duchy strategy ends the game with \(3\) more Duchies than Dukes.
With \(p\) $5-purchases, this yields: \begin{align} V & = \frac{(p + 4)(p+2)}{4} \text{ for }p\text{ even,} \\ V &= \frac{(p+3)^2}{4} \text{ for }p\text{ odd.} \end{align}
So, to beat a player who has \(4\) Provinces, the ideal strategy needs \(p=7\) $5-purchases (\(5\) Duchies and \(2\) Dukes).
To beat a player who has \(8\) Provinces, the strategy needs \(p=11\) $5-purchases (\(7\) Duchies and \(4\) Dukes).